Natural Science Forum / Physics / Relativity / January 2007

No performing involved.  The trace of the Einstein tensor is the same
as the trace of the Ricci tensor, except with the opposite sign.

Yes.
Eliminate the 1/2, that's a non-zero constant.
Eliminate R or the result is trivial, so all that is left is g_uv = 0.
But we knew that, everything about relativity is built on zero.
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
c = 0/0

Hmm. I'm not sure what is meant by "reverse trace", but I do
understand the conclusion:

Start with
   R_uv = 1/2 R g_uv

multiplying both sides by g^uv and summing gives

   sum over u,v of R_uv g^uv = sum over u,v 1/2 R g_uv g^uv

The left side gives, by definition, R.
The right side gives 2R (because sum over u,v  g_uv g^uv = 4).

so we have

   R = 2R

which implies R=0.

--
Daryl McCullough
Ithaca, NY